What is the solution for y^{''}=4y,y(0)=1,y(1)=0 ?

The solution for y^{''}=4y,y(0)=1,y(1)=0 is y=(1-(e^4)/(e^4-1))e^{2t}+(e^4)/(e^4-1)e^{-2t}

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y^{''}=4y,y(0)=1,y(1)=0

y^{^{''}} = 4 y, y (0) = 1, y (1) = 0

y = (1 - \frac{e ^{4}}{e ^{4} - 1}) e^{2 t} + \frac{e ^{4}}{e ^{4} - 1} e^{- 2 t}

Solution steps

y^{''} (t) = 4 y

Solve linear ODE:

y = (1 - \frac{e ^{4}}{e ^{4} - 1}) e^{2 t} + \frac{e ^{4}}{e ^{4} - 1} e^{- 2 t}

y = (1 - \frac{e ^{4}}{e ^{4} - 1}) e^{2 t} + \frac{e ^{4}}{e ^{4} - 1} e^{- 2 t}

y^{^{''}} - 4 a^{2} y^{^{'}} + y = 0

y^{^{''}} + 36 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} + 8 y^{^{'}} = 0, y (0) = 3, y^{^{'}} (0) = 16

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{3}{2}

\frac{1}{2} y^{^{''}} - \frac{1}{3} y^{^{'}} - 2 y = 0

What is the solution for y^{''}=4y,y(0)=1,y(1)=0 ?
The solution for y^{''}=4y,y(0)=1,y(1)=0 is y=(1-(e^4)/(e^4-1))e^{2t}+(e^4)/(e^4-1)e^{-2t}